Six mobsters have arrived at the theater for the premiere of

[aps_asks]

Hi Tim ,

I am not following the explanations provided by MGMAT. Rarely , i do feel that MGMAT does not provide good explanations in their answers and this problem is one of them.

I am following the explanation provided by GMAT 2007

[price.karr]

Hi Tim/Stacy,

Assuming that JF will be in front or behind one another in exactly half of the permutations makes sense, but does this still hold true if there were an odd number of people? ie, Would 5!/2 be the correct answer if there were only 5 people in line?

[jnelson0612]

Hi Tim ,

I am not following the explanations provided by MGMAT. Rarely , i do feel that MGMAT does not provide good explanations in their answers and this problem is one of them.

I am following the explanation provided by GMAT 2007

aps, what are you not understanding about the explanation I just gave you? Please give us more info.

[jnelson0612]

Hi Tim/Stacy,

Assuming that JF will be in front or behind one another in exactly half of the permutations makes sense, but does this still hold true if there were an odd number of people? ie, Would 5!/2 be the correct answer if there were only 5 people in line?

Good question! Let's look at that using an even easier example of 3 people. Let's say that we still have Frankie and Joey, but they are joined by one other mobster, Anthony. Let's first look at the way they can arrange themselves:

1)AFJ
2)AJF
3)FAJ
4)FJA
5)JAF
6)JFA

They can arrange themselves 6 ways, or 3! (3*2*1).

Now, let's look further. In arrangements #1, 3, and 4 Frankie is ahead of Joey; in arrangements #2, 5, and 6 Joey is ahead of Frankie. Once again, each man is in front of the other half the time.

I like your question because it helps me drive home the point that the number of other people doesn't matter; they really are just filler around Frankie and Joey. Frankie will be in front half the time; Joey will be in front half the time Thus, whether the number of mobsters is odd or even is irrelevant. Thanks for asking! :-)

[rudransh]

The below approach is from MGMAT 1st edition

to begin with

1) J can occupy any of the 6 positions = 6
2) other 4 can occupy the remaining 4 positions in any order = 4!
3) Now for F (who wants to stand anywhere behind J)
3.1) If J takes 1st position then F has 5 out of 6 positions = 5/6
3.2) If J takes 2nd position then F has 4 out of 6 positions = 4/6
3.3) If J takes 3rd position then F has 3 out of 6 positions = 3/6
3.4) If J takes 4th position then F has 2 out of 6 positions = 2/6
3.5) If J takes 5th position then F has 1 out of 6 positions = 1/6
3.6) If J takes 6th position then F has 0 out of 6 positions = 0/6
add (5/6 + 4/6 + 3/6 + 2/6 + 1/6 + 0) = 15/6 for F

Total arrangements = 6 * 4! * 15/6 = 360

The above works well for most of combinations questions with restriction

For some reason, this approach was there in 1st edition of MGMAT books but I could not find this in 4th edition...not sure why?

[jnelson0612]

The below approach is from MGMAT 1st edition

to begin with

1) J can occupy any of the 6 positions = 6
2) other 4 can occupy the remaining 4 positions in any order = 4!
3) Now for F (who wants to stand anywhere behind J)
3.1) If J takes 1st position then F has 5 out of 6 positions = 5/6
3.2) If J takes 2nd position then F has 4 out of 6 positions = 4/6
3.3) If J takes 3rd position then F has 3 out of 6 positions = 3/6
3.4) If J takes 4th position then F has 2 out of 6 positions = 2/6
3.5) If J takes 5th position then F has 1 out of 6 positions = 1/6
3.6) If J takes 6th position then F has 0 out of 6 positions = 0/6
add (5/6 + 4/6 + 3/6 + 2/6 + 1/6 + 0) = 15/6 for F

Total arrangements = 6 * 4! * 15/6 = 360

The above works well for most of combinations questions with restriction

For some reason, this approach was there in 1st edition of MGMAT books but I could not find this in 4th edition...not sure why?

Well, in my opinion, this obviously is one way to do the problem and an interesting one at that. However, this kind of calculation is not easy for most people and will be difficult to execute in two minutes. Since we know that time is the number one reason people score below their ability levels on the test, we need to equip them with time efficient ways to do problems. Because of this, I'd rather my students think conceptually about this problem, realizing that the other mobsters are filler and in half the orders F will be in front of J and in half the orders J will be in front of F. I'd much rather people understand this concept than go through this long process. But that's just me!

[temizmustafa]

Dear Instructors and Gmat takers,

I considered JF or FJ like a one person and used 5!=120 and multiplied by 2. and found 240 which is not even in choices. Why cant we consider FJ or JF as one person then multiply by 2..?

There 5 ways that JF or FJ can be together and 4! of arranging the others 5! = 120 and *2 = 240???

Thank you in advance.

[tim]

Jamie already answered this on June 25, 2011. you are trying to make the two people adjacent, which has nothing to do with this problem..

[vrampall12]

I got a similar problem on a practice exam:

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?

Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him

the bolded part seems to imply the Frankie CANNOT be directly behind so XXXXJF is not an option, but the answer is still 360.

Am I missing something?

[tim]

Your interpretation of the bold part is incorrect. It is absolutely false to claim that "not necessarily" means the same thing as "not possibly".

[sg2010]

Can someone reply to the most recent comment, about explaining how to add up the various scenarios? i.e. the 5! - 2*4!.?

[Sage Pearce-Higgins]

I don't understand which comment you're referring to. Please could you make your question clearer?

[JackF968]

I'm confused how this is being missed. Frankie wants to be behind Joey so the order (being read from a computer) will be left-to-right or right-to-left. Frankie has to have Joey in front of him, so direction matters. That means Joey cannot be the last in line because that would mean Joey is behind Frankie. Joey also cannot be 2nd to last in line because then for Frankie to be behind him, they would have to be next to eachother. That means it would only work with Joey in the 1st, 2nd, 3rd, or 4th position and NOT in the 5th or 6th position. Wouldn't that be

6!/ (4!2!) ?

Please help!

[Sage Pearce-Higgins]

Joey also cannot be 2nd to last in line because then for Frankie to be behind him, they would have to be next to each other.

Please check where you got this piece of information from. I can't find anything in the question about them not standing next to each other. Also, please could you explain how your understanding related to your calculation? Where does the 4!2! come from in your formula?

[JackF968]

"Six mobsters have arrived at the theater for the premiere of the film “Goodbuddies.” One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?"

This is from my CAT Exam Review:
"...insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him."

I interpreted that as Frankie can't be in position 1 because then Joey would be behind him.
He can't be in position 2 if Joey is the one in front of him
He can only be in Position 3 if Joey is in Position 1, meaning Joey is both in front of him and Frankie is not right behind him
He can only be in Position 4 if Joey is in Position 1 or 2, meaning Joey is both in front of him and Frankie is not right behind him

.... etc etc

Not that the permutations are correct way to think about it mathematically but that led me to him being able to be in four spots (4!) but not in two spots (2!), out of six spots (numerator)... wrong or right. The setup I described above is how I think about the situation... math formula aside.

And thank you for responding!